Question

Calculate the entropy change when 36 g of ice is heated at standard pressure from 230 K to 320 K. Take the molar heat capacities at constant pressure, Cp,m, of water and ice to be
75.3 and 37.7 J mol 1 -1 respectively, and the molar enthalpy of fusion of ice to be 6.02 kJ mol-^. The molar mass of water is 18.0 g mol-1, and ice melts at a temperature of 273.15 K

Answer 1

$$36 \mathrm{g} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} 230 \mathrm{K} \rightarrow^{1} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} 273.15 \mathrm{K} \rightarrow^{2} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} 273.15 \mathrm{K} \rightarrow^{3} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} 320 \mathrm{K}$$

Amount of ice $\mathrm{n} = \frac{36g}{18g/mol} = 2\mathrm{mol}$

$$\begin{array}{l} \Delta \mathrm{S}_{1}^{0} = n \mathrm{C}_{p} \ln \frac{T_{f}}{T_{i}} = 2 * 0.0377 * \ln \frac{273.15}{230} = 0.013 \mathrm{kJ/K} \\ \Delta \mathrm{S}_{2}^{0} = \frac{n \cdot \Delta H_{fus}^{0}}{T} = \frac{2 * 6.02}{273.15} = 0.044 \mathrm{kJ/K} \\ \Delta \mathrm{S}_{3}^{0} = n \mathrm{C}_{p} \ln \frac{T_{f}}{T_{i}} = 2 * 0.0753 * \ln \frac{320}{273.15} = 0.024 \mathrm{kJ/K} \\ \Delta \mathrm{S}_{\text{total}}^{0} = \Delta \mathrm{S}_{1}^{0} + \Delta \mathrm{S}_{2}^{0} + \Delta \mathrm{S}_{3}^{0} = 0.081 \mathrm{kJ/K} = 81 \mathrm{J/K} \\ \end{array}$$

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